Question

A block of a plastic material floats in water with 35.6% of its volume under water. What is the density of the block in kg/m3

Answer 1

Answer:

The  density is  $\rho_{block} = 356 \ kg/m^3$

Explanation:

From the question we are told that

   The proportion of the plastic under water is  p =  0.356

Generally at equilibrium the buoyant force is equal to the weight of the block i.e

         $B = mg$

Generally the Buoyant force is mathematically represented as

          $B = \rho * V * g$

Here $\rho$ density of the water with value

$\rho = 1000 kg/m^3$ ,

V is the volume of water displaced by the block which is $p * V_{block$ ,  

So

    $1000 * 0.356 V_{block} * g = m * g$

Here m is the mass of the block which is mathematically represented  as

        $m = \rho_{block} * V_{block}$

So

         $1000 * 0.356 V_{block} * g = \rho_{block} * V_{block}* g$

         $1000 * 0.356 V_{block} = \rho_{block} * V_{block}* g$

       =>    $\rho_{block} = 1000 * 0.356$

      =>    $\rho_{block} = 356 \ kg/m^3$